- #1

Vitor Pimenta

- 10

- 1

## Homework Statement

A particle of mass m moves in a "central potential" , V(r), where r denotes the radial displacement of the particle from a fixed origin.

From Hamilton´s equations, obtain a "one-dimensional" equation for [tex]{\dot p_r}[/tex], in the form [tex]{{\dot p}_r} = - \frac{\partial }{{\partial r}}\left[ {{V_{eff}}\left( r \right)} \right][/tex], where [tex]{V_{eff}}\left( r \right)[/tex] denotes an "effective" potential that is a funcion of r only.

## Homework Equations

Hamiltonian: [tex]H = \frac{{{p_r}^2}}{{2m}} + \frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)[/tex] , where L is the angular momentum with respect to the origin, which is a constant of the motion.

[tex]\frac{{\partial H}}{{\partial r}} = - {\dot p_r}[/tex]

## The Attempt at a Solution

[tex]\begin{array}{l}

{{\dot p}_r} = - \frac{{\partial H}}{{\partial r}} = - \frac{\partial }{{\partial r}}\left[ {\frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)} \right]\\

\therefore {V_{eff}}\left( r \right) = \frac{{{L^2}}}{{2m{r^2}}} + V\left( r \right)

\end{array}[/tex]

The problem is that it doesn´t make sense to me that the effective potential is different than the normal one ( V(r) ). Besides, the force acting on the particle shouldn´t have a dependence on its angular momentum L.